使用定义为函数的上下文管理器,可以很容易地以编程方式从一个单独的(或递归的)上下文管理器中输入,如下所示:
@contextmanager
def enter(times):
if times:
with enter(times - 1) as tup:
print 'entering {}'.format(times)
yield tup + (times,)
print 'exiting {}'.format(times)
else:
yield ()
运行这个:
In [11]: with enter(4) as x:
....: print x
....:
entering 1
entering 2
entering 3
(1, 2, 3)
exiting 3
exiting 2
exiting 1
进出账记账都给你搞定了,多好啊!但是如果你有一个类而不是一个函数呢?
class Enter(object):
def __init__(self, times):
self.times = times
def __enter__(self):
print 'entering {}'.format(self.times)
if self.times:
with Enter(self.times - 1) as tup: # WRONG
return tup + (self.times,)
return ()
def __exit__(self, *_):
print 'exiting {}'.format(self.times)
运行这个是错误的,因为你在运行 with-block 中的任何代码之前进入和退出嵌套调用:
In [12]: with Enter(3) as tup:
print tup
....:
entering 3
entering 2
entering 1
entering 0
exiting 0
exiting 1
exiting 2
(1, 2, 3)
exiting 3
规定:强制客户端自己使用
ExitStack
是 Not Acceptable ;内部调用必须像在生成器中一样被封装。涉及 Enter
维护其自己的私有(private)堆栈的解决方案也是次优的(在现实生活中,有必要以线程安全的方式将内部 __exit__
调用与内部 __enter__
调用相匹配,但我想避免那种即使在这个简单的例子中,也尽可能多地手动记账。) 最佳答案
在 __enter__
中使用嵌套的上下文管理器似乎很神奇。
看一下这个:
class Enter(object):
def __init__(self, times):
self.times = times
def __enter__(self):
print('entering {}'.format(self.times))
if self.times:
with Enter(self.times - 1) as tup: # WRONG
print('returning {}'.format(tup))
return tup + (self.times,)
print('returning () from times={}'.format(self.times))
return ()
def __exit__(self, *_):
print('exiting {}'.format(self.times))
with Enter(3) as tup:
print(tup)
运行此打印
entering 3
entering 2
entering 1
entering 0
returning () from times=0
returning ()
exiting 0
returning (1,)
exiting 1
returning (1, 2)
exiting 2
(1, 2, 3)
exiting 3
我认为这在某种程度上是有道理的。心智模型可能是,当您调用
with Enter(3) ...
时,必须“完成” __enter__
方法,“完成”意味着进入和退出所有上下文管理器。def foo():
with Enter(2) as tup:
return tup
# we expect Enter to exit before we return, so why would it be different when
# we rename foo to __enter__?
让我们明确地做到这一点。
In [3]: %paste
class Enter(object):
def __init__(self, times):
self.times = times
self._ctx = None
def __enter__(self):
print('entering {}'.format(self.times))
if self.times:
self._ctx = Enter(self.times - 1)
tup = self._ctx.__enter__()
return tup + (self.times,)
else:
return ()
def __exit__(self, *_):
if self._ctx is not None:
self._ctx.__exit__()
print('exiting {}'.format(self.times))
In [4]: with Enter(3) as tup:
...: print(tup)
...:
entering 3
entering 2
entering 1
entering 0
(1, 2, 3)
exiting 0
exiting 1
exiting 2
exiting 3
(在@jasonharper 的指导下回答。)
关于python - 在 __enter__ 中输入上下文管理器,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/47705573/